How to prove $M\in O^+(n,1)$ can always be decomposed to the product of two matrices To be specific, I want to prove that for any $M\in \mathbb{R}_{(n+1)\times (n+1)} $ (See update below)
$$
M^T
\begin{bmatrix}
I_n & 0\\
0 & -1
\end{bmatrix}
M=
\begin{bmatrix}
I_n & 0\\
0 & -1
\end{bmatrix}
$$
there exists $Q$ and $B$ that
$$
Q=
\begin{bmatrix}
Q_0 & 0\\
0 & 1
\end{bmatrix},\quad \text{where }Q_0\in O(n)\\
B=
\begin{bmatrix}
I_{n-1} & 0 & 0\\
0 & \cosh t & \sinh t \\
0 & \sinh t & \cosh t
\end{bmatrix}
$$
satisfying $M=QB$.
I don't have any idea. I can only think of writing $M$ in the form of a block matrix.
$$
M=
\begin{bmatrix}
M_0 & \alpha\\
\beta ^T & x
\end{bmatrix}
$$
substitute it into the first equation will come to
$$
M_0^T\alpha=x\beta\\
\alpha^T\alpha-x^2=-1\\
M_0^TM_0-\beta \beta^T=I_n
$$
I don't know how to introduce $Q_0$, not to say B.
Any help please?

Update: Sorry for misstating. $M$ should be $M\in O^+(n,1)$, the subgroup of isometries that are time orientation-preserving (physicists
call it the orthochronous Lorentz group), and $B$ should be any matrix of a Lorentz boost.
 A: The statement is false. The bottom row of your $QB$ has the sign pattern $(0,\ldots,0,\ast,+)$, but this is not necessarily the case for $M$ when $M=-I$ or $$
M=\pmatrix{\cosh t&&\sinh t\\ &I_{n-1}\\ \sinh t&&\cosh t},\ n\ge1,\,t\ne0.
$$
It is true, however, that $M=(Q_1\oplus\pm1)B(Q_2\oplus1)$ for some $Q_1,Q_2\in O(n)$ and some $B$ that takes the form as specified in your question.
Let $M=UP$ be the polar decomposition of $M$, where $U$ is orthogonal and $P$ is positive definite. Let $D=I_n\oplus-1$. From $M^TDM=D$ we infer that
\begin{align}
PU^TDUP&=D,\\
DUP&=UP^{-1}D,\\
(DUP)^TDUP&=(UP^{-1}D)^TUP^{-1}D,\\
P^2&=DP^{-2}D=(DP^{-1}D)^2,\\
P&=DP^{-1}D,\\
\end{align}
where the last line is due to the fact that every positive definite matrix has a unique positive definite square root. Now partition $P$ as $\pmatrix{A&b\\ b^T&c}$ where $A\succ0$ and $c>0$. Then
$$
I=DPDP=\pmatrix{A&b\\ -b^T&-c}\pmatrix{A&b\\ -b^T&-c}=\pmatrix{A^2-bb^T&(A-cI)b\\ b^T(A-cI)&c^2-b^Tb}.
$$
Therefore $c\ge1,\ b=(c^2-1)^{1/2}u$ for some unit eigenvector $u$ of $A$ corresponding to the eigenvalue $c$, and $A^2=I+(c^2-1)uu^T$. Hence $A=I+(c-1)uu^T$ and
$$
P=\pmatrix{I+(c-1)uu^T&(c^2-1)^{1/2}u\\ (c^2-1)^{1/2}u^T&c}
=\pmatrix{Q_2^T\\ &1}\underbrace{\pmatrix{I_{n-1}\\ &\cosh t&\sinh t\\ &\sinh t&\cosh t}}_B\pmatrix{Q_2\\ &1}
$$
where $c=\cosh t$ and $Q_2\in O(n)$ is any orthogonal matrix whose last row is $u^T$.
It follows that $M=QB(Q_2\oplus1)$, where $Q=U(Q_2^T\oplus1)$. So, from $M^TDM=D$, we obtain $Q^TDQ=B^{-1}(Q_2\oplus1)D(Q_2^T\oplus1)B^{-1}=B^{-1}DB^{-1}=D$. Hence $DQ=QD$ and $Q=Q_1\oplus\pm1$ for some orthogonal matrix $Q_1$. Now we are done.
Edit. If you also require $M$ to be a member of the orthochronous group (i.e. $M^TDM=D,\,\det(M)=1$ and $M_{n+1,n+1}\ge1$), then the $Q$ above must take the form of $Q_1\oplus1$ (rather than $Q_1\oplus\pm1$) with $\det(Q_1)=\det(Q_2)$. The matrices $Q_1$ and $Q_2$ can be chosen from $SO(n)$ if $t$ is allowed to be negative.
